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kari74 [83]
3 years ago
7

What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?

Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
6 0

Answer: The volume of 0.640 grams of O_{2} gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of O_{2} gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of O_{2} (molar mass = 32.0 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L

Thus, we can conclude that the volume of 0.640 grams of O_{2} gas at Standard Temperature and Pressure (STP) is 0.449 L.

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you fill a rigid steel container that has a volume of 20 L with nitrogen gas to a final pressure of 2 x 10^4 kpa at 23 Celsius.
garik1379 [7]

Answer:

4.549 kg.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 2 x 10⁴ kPa/101.325 = 197.4 atm).

V is the volume of the gas in L (V = 20.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 23° C + 273 = 296 K).

<em>∴ n = PV/RT =</em> (197.4 atm)(20.0 L)/(0.0821 L.atm/mol.K)(296 K) = <em>162.5 mol.</em>

  • To find the mass of N₂ in the cylinder, we can use the relation:

<em>mass of N₂ = (no. of moles of N₂)*(molar mass of N₂) = </em>(162.5 mol)*(28.0 g/mol) = <em>4549 g = 4.549 kg.</em>

3 0
3 years ago
PLEASE HELP ASAP!!!!
Vsevolod [243]

Answer:

The answer to your question is  C = 0.037 cal/g°C

Explanation:

Data

mass of the wire = m = 237 g

temperature 1 = T1 = 25°C

temperature 2 = T2 = 107°C

Heat = Q = 722 cal

Specific heat = C

Process

1.- Write the formula to find the specific heat

            Q = mC(T2 - T1)

-Solve for C

            C = Q / m(T2 - T1)

2.- Substitution

             C = 722 / 237(107 - 25)

3.- Simplification

              C = 722 / 237(82)

              C = 722 / 19434

4.- Result

               C = 0.037 cal/g°C

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Answer:

I think they both need heat.

Explanation:

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